\title{Regime 1 Algebra}
\author{Xinya Zhang}
\date{\today}

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\begin{document}
\maketitle
\section{Hamiltonian System}
For the easy-read purpose, I copy all the first order equations and differential equations of co-state variables here, and I will use the equation numbers listed below from now on:
\begin{eqnarray}
\frac{\partial \mathcal{H}}{\partial c} &=& c^{-\gamma} -\lambda = 0
\label{eq:FOCc}  \\
\frac{\partial \mathcal{H}}{\partial i} &=& -\lambda + q = 0 \label{eq:FOCi}\\
\frac{\partial \mathcal{H}}{\partial j} &=& -\lambda + \eta\psi B + \mu = 0; \mu j=0, \mu\ge 0, j \ge 0 \label{eq:FOCj}\\
\frac{\partial \mathcal{H}}{\partial n} &=& -\lambda +\nu+\omega = 0, \omega n=0, \omega\ge 0, n\ge 0  \label{eq:FOCn}\\
\frac{\partial \mathcal{H}}{\partial R} &=& -\lambda g(S,N)+\epsilon+\sigma Q+\xi = 0, \xi R=0, \xi\ge 0, R\ge 0 \label{eq:FOCR} \\
\frac{\partial \mathcal{H}}{\partial B} &=& -\lambda(\Gamma_1+H)^{-\alpha}+\epsilon+\eta(1+\psi j)+\zeta=0, \zeta B=0, \zeta\ge 0, B\ge 0 \label{eq:FOCB}
\end{eqnarray}
The differential equations:
\begin{eqnarray}
\dot{q} &=& \beta q-\frac{\partial \mathcal{H}}{\partial k} = (\beta+\delta) q - \lambda A+\epsilon A \label{eq:qdot}\\
\dot{\eta} &=& \beta \eta- \frac{\partial \mathcal{H}}{\partial H} =\beta\eta - \lambda\alpha(\Gamma_1+H)^{-\alpha-1}B \label{eq:etadot} \\
\dot{\sigma} &=& \beta \sigma - \frac{\partial \mathcal{H}}{\partial S} = \beta \sigma +\lambda \frac{\partial g}{\partial S}R  \label{eq:sigmadot} \\
\dot{\nu} &=& \beta \nu - \frac{\partial \mathcal{H}}{\partial N} = \beta\nu +\lambda \frac{\partial g}{\partial N}R  \label{eq:nudot}
\end{eqnarray}

I also put the budget constraint and the other useful equations here:
\begin{eqnarray}
 y &=& c+i+j+n+g(S,N)R+pB \label{eq:Budget} \\
 p &=& (\Gamma_1+H)^{-\alpha},\:H<\Gamma_2^{-1/\alpha}-\Gamma_1 \label{eq:RenewCost} \\
\dot{k} &=& i-\delta{k} \label{eq:kdot} \\
\dot{H} &=& B(1+\psi j),\: H<\Gamma_2^{-1/\alpha}-\Gamma_1 \label{eq:Hdot}\\
\dot{N} &=& n \label{eq:Ndot}\\
\dot{S} &=& QR \label{eq:Sdot}\\
y &=& Ak \label{eq:prod}
\end{eqnarray}




\section {Regime 1 without Tax}

in the Regime $[0,T_1]$, $R>0$, $n>0$, $B=0$, and $j=0$. Since $R>0$, we have $\xi$=0, and then, from FOC condition(5),we get:
\begin{equation}
\epsilon=\lambda g(S,N)-\sigma Q \label{eq:epsilon1}
\end{equation}
Since $n>0$, we know $\omega=0$. According to FOC condition (2) and (4), we find $\lambda=q=\nu$, and hence $\dot{\lambda}=\dot{q}=\dot{\nu}$. From equation \eqref{eq:qdot}, \eqref{eq:nudot} and \eqref{eq:epsilon1},we have:
\begin{equation}
[\delta+g(S,N)A-\frac{\partial g}{\partial N}R-A]\lambda = \sigma QA
\label{eq:qdoteqnudot}
\end{equation}

From $\dot{\lambda}=\dot{q}=\dot{\nu}$, and equation (10), we obtain the differential equation of $\lambda$:
\begin{equation}
\dot{\lambda} = \lambda(\beta+\frac{\partial g}{\partial N}Ak) \label{eq:q_nu_lambda}
\end{equation}

Now we rewrite equation (\ref{eq:qdoteqnudot}) for easy differentiation purpose (Move $\lambda$ to the right hand side and divide A on both sides):

\begin{equation}
\frac{\delta}{A}+g(S,N)-\frac{\partial g}{\partial N}k-1 = \frac{\sigma Q}{\lambda}
\label{eq:qdoteqnudot'} \tag{\ref{eq:qdoteqnudot}'}
\end{equation}

Differentiate (\ref{eq:qdoteqnudot'}) on both sides with time $t$, we have the equations below step by step:
\begin{eqnarray}
\frac{\mathrm{d}\mathbf{LHS}}{\mathrm{d}{t}}
&\Longrightarrow&
\frac{\partial g}{\partial S}\dot{S}+\frac{\partial g}{\partial N}\dot{N}-
	\left(\frac{\partial^2g}{\partial S\partial N}\dot{S}+\frac{\partial^2g}{\partial N^2}\dot{N}\right)k-
	\frac{\partial g}{\partial N}\dot{k} \nonumber\\
&\stackrel{(\ref{eq:Sdot})\,(\ref{eq:Ndot})\,(\ref{eq:kdot})}{\Longrightarrow}&
\frac{\partial g}{\partial S}QAk+\frac{\partial g}{\partial N}n-
	\left(\frac{\partial^2g}{\partial S\partial N}QAk+\frac{\partial^2g}{\partial N^2}n\right)k-
	\frac{\partial g}{\partial N}(i-\delta k) \nonumber\\
\frac{\mathrm{d}\mathbf{RHS}}{\mathrm{d}{t}}
&\Longrightarrow&
 \frac{(\dot{\sigma}Q+\sigma\dot{Q})\lambda-\sigma Q\dot{\lambda}}{\lambda^2} \nonumber \\
&\stackrel{(\ref{eq:sigmadot})\,(\ref{eq:q_nu_lambda})}{\Longrightarrow}&
\frac{(\beta\sigma+\lambda\frac{\partial g}{\partial S}Ak)Q+\sigma\pi Q-\sigma Q (\beta+\frac{\partial g}{\partial N}Ak)}{\lambda}  \nonumber\\
&\Longrightarrow&
\frac{\partial g}{\partial S}QAk + \frac{\sigma\pi Q-\frac{\partial g}{\partial N}\sigma QAk}{\lambda} \nonumber
\end{eqnarray}

Let $\frac{\mathrm{d}\mathbf{LHS}}{\mathrm{d}{t}} = \frac{\mathrm{d}\mathbf{RHS}}{\mathrm{d}{t}}$ and simplify the equation, we can obtain the final expression:
\begin{equation}
\lambda\bigg[\frac{\partial g}{\partial N}\bigg(n+\delta k+\frac{\sigma QAk}{\lambda}-i\bigg)-\frac{\partial^2g}{\partial S\partial N}QAk^{2}-\frac{\partial^2g}{\partial N^2}nk\bigg]=\sigma\pi Q  \label{eq:diff1}
\end{equation}

A second relationship between $i$ and $n$ is given by the budget constraint\eqref{eq:Budget}. We rewrite it here:
\begin{equation}
i=Ak(1-g(S,N))-\lambda^{-1/\lambda}-n \label{eq:i_fossregime}
\end{equation}

Substituting\eqref{eq:i_fossregime} into \eqref{eq:diff1}, we can obtain and equation to solve n:
 \begin{equation}
\begin{split}
& n\lambda\left(\frac{\partial ^{2}g}{\partial N^{2}}k-2\frac{\partial g}{%
\partial N}\right) =\\&\lambda \left[ \frac{\partial g}{\partial N}\bigg[k(\delta
+g(S,N)A-A+\frac{\sigma QA}{\lambda })+\lambda ^{-1/\gamma }\bigg]-\frac{\partial
^{2}g}{\partial S\partial N}QAk^{2}\right] -\sigma \pi Q
\end{split}
\label{eq:n_soln}
\end{equation}




\section{Regime 1 with Tax on $n$}

Introducing taxes $\tau_n$ on $n$ during Regime 1, the budget constraint in that
regime becomes:

\begin{equation}
c+i+n(1+\tau _{n})+g(S,N)R=y+T \label{eq:Budget_tau_n}
\end{equation}
with a corresponding budget constraint fr the government given by $T=\tau_{n}n$.

The first order conditions remain the same as before except one:
\begin{equation}
\frac{\partial \mathcal{H}}{\partial n}=-\lambda (1+\tau _{n})+\nu +\omega
=0,\omega n=0,\omega \geq 0,n\geq 0 \label{eq:FOCn'} \tag{\ref{eq:FOCn}'}
\end{equation}


We also assume parameter values and taxes are chosen so that investment in
fossil fuel technology is productive, that is, $n>0$. Then $\omega =0$. According to FOC condition \eqref{eq:FOCi} and \eqref{eq:FOCn'}, we find $\lambda=q=\frac{\nu}{(1+\tau _{n})}$, and hence $\dot{\lambda}=\dot{q}=\frac{\dot{\nu}}{(1+\tau _{n})}$. From equation \eqref{eq:FOCn'}, \eqref{eq:qdot}, \eqref{eq:nudot} and \eqref{eq:epsilon1},we have:
\begin{eqnarray}
\frac{\delta }{A}+g(S,N)-\frac{1}{(1+\tau _{n})}\frac{\partial g}{%
\partial N}k-1 = \frac{\sigma Q}{\lambda} \label{eq:qdoteqnudot_tau} \\
\dot{\lambda} = \lambda\bigg(\beta+\frac{1}{1+\tau_n}\frac{\partial g}{\partial N}Ak\bigg) \label{eq:q_nu_lambda_tau}
\end{eqnarray}


Compare \eqref{eq:qdoteqnudot_tau} to equation \eqref{eq:qdoteqnudot'}, the effect of $N$ on cost $g$ is weaken by $\frac{1}{1+\tau_n}$. Now we differentiate \eqref{eq:qdoteqnudot_tau} on both sides with time $t$, step by step:

 \begin{eqnarray}
\frac{\mathrm{d}\mathbf{LHS}}{\mathrm{d}{t}}
&\Longrightarrow&
\frac{\partial g}{\partial S}\dot{S}+\frac{\partial g}{\partial N}\dot{N}-
	\frac{1}{1+\tau_n}(\frac{\partial^2g}{\partial S\partial N}\dot{S}+\frac{\partial^2g}{\partial N^2}\dot{N})k-
	\frac{1}{1+\tau_n}\frac{\partial g}{\partial N}\dot{k} \nonumber\\
&\Longrightarrow&
\frac{\partial g}{\partial S}QAk+\frac{\partial g}{\partial N}n-\frac{1}{1+\tau_n}
	\left(\frac{\partial^2g}{\partial S\partial N}QAk+\frac{\partial^2g}{\partial N^2}n\right)k-
	\frac{1}{1+\tau_n}\frac{\partial g}{\partial N}(i-\delta k) \nonumber\\
\frac{\mathrm{d}\mathbf{RHS}}{\mathrm{d}{t}}
&\Longrightarrow&
 \frac{(\dot{\sigma}Q+\sigma\dot{Q})\lambda-\sigma Q\dot{\lambda}}{\lambda^2} \nonumber \\
&\stackrel{(\ref{eq:sigmadot})\,(\ref{eq:q_nu_lambda_tau})}{\Longrightarrow}&
\frac{(\beta\sigma+\lambda\frac{\partial g}{\partial S}Ak)Q+\sigma\pi Q-\sigma Q(\beta+\frac{1}{1+\tau_n}\frac{\partial g}{\partial N}Ak)}{\lambda}  \nonumber\\
&\Longrightarrow&
\frac{\partial g}{\partial S}QAk + \frac{\sigma\pi Q-\frac{1}{1+\tau_n}\frac{\partial g}{\partial N}\sigma QAk}{\lambda} \nonumber
\end{eqnarray}

Similarly, $\frac{\mathrm{d}\mathbf{LHS}}{\mathrm{d}{t}} = \frac{\mathrm{d}\mathbf{RHS}}{\mathrm{d}{t}}$ implies:
\begin{equation}
\frac{\partial g}{\partial N}\bigg(n+\frac{1}{1+\tau_n}(\delta k+\frac{\sigma QAk}{\lambda}-i)\bigg)-\frac{1}{1+\tau_n}\bigg(\frac{\partial^2g}{\partial S\partial N}QAk+\frac{\partial^2g}{\partial N^2}n\bigg)k=\frac{\sigma\pi Q}{\lambda}  \label{eq:diff1_tau_n}
\end{equation}

A second relationship between $i$ and $n$ is given by the budget constraint\eqref{eq:Budget_tau_n}. Note that the expression is the same as \eqref{eq:i_fossregime} in the zero tax regime.
Combine these two, we have an equation to solve n:
\begin{equation}
\begin{split}
& n\left(\frac{1}{1+\tau_{n}}\frac{\partial ^{2}g}{\partial N^{2}}k-(1+\frac{1}{1+\tau_{n}})\frac{\partial g}{%
\partial N}\right) =\\&\frac{1}{1+\tau_{n}} \left[ \frac{\partial g}{\partial N}\bigg[k(\delta
+g(S,N)A-A+\frac{\sigma QA}{\lambda })+\lambda ^{-1/\gamma }\bigg]-\frac{\partial
^{2}g}{\partial S\partial N}QAk^{2}\right] -\frac{\sigma \pi Q}{\lambda}
\end{split}
\label{eq:n_soln_tau_n}
\end{equation}
Simplify it:
\begin{equation}
n=\frac{\frac{\partial g}{\partial N}[k(\delta
+g(S,N)A-A+\frac{\sigma QA}{\lambda })+\lambda ^{-1/\gamma }]-\frac{\partial^{2}g}{\partial S\partial N}QAk^{2}-\frac{\sigma \pi Q(1+\tau_{n})}{\lambda} }{\frac{\partial ^{2}g}{\partial N^{2}}k-(2+\tau_{n})\frac{\partial g}{
\partial N}}
\end{equation}


\section{Subsidy on j}



\end{document}
